First read example 4.6.5. from the Damiano and Little text.

From this we see that the conic section is a hyperbola. (See Figure 4.7.) In (x1, x2)-coordinates, the vertices of the hyperbola are at the points (3/V/50, 1/V/50) and (-3/V50, -1/V50). The asymptotes are the lines x, = +.x2 [or x, = -2x2 and x1 = (1/2).x2]. Note that as we expect, the picture shows clearly that the new ( ). and *;) axes are simply rotated from the standard X1 = 12X2 X1 = -2X: Figure 4.73 34 1 into the graph of 5x2 5y2 = 1, which is more easily identifiable as an hyperbola. To see this, rename In this example, the authors diagonalize the matrix ( ) to turn the graph of 4x2 + 6xy 4y2 = the variables from the textbook "x" and \"y". The question: Explain in detail how one can turn ax2 + bey ay2 = 1 into me2 ty2 = 1 for some real numbers r a b and I by diagonalizing a matrix of the form ( b ) . Note that a and b are real strictly positive a numbers. The explanation in example 4.6.5 is missing some details. Be sure to fill in the blanks with any theorems that are being used or details that are missing. You do not have to construct the matrix Q or find any eigenvectors. However, you should still find the eigenvalues in terms of a and b. Then explain how you could find matrix Q (without actually finding it), explain how you would use it, and discuss how to get the equation of the hyperbola in the new basis. Hint 1: You are welcome to look up hyperbolas online. However, knowledge beyond the general equation of an hyperbola is not required for this question. Hint 2: The textbook writes a symbol that looks like Q' but is actually Qr , the transpose of